A rigid wire loop of square shape having side of length L and resistance R is moving along the X-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive. Which of the following schematic plot(s) is (are) correct? (Ignore gravity)

Electromagnetic Induction PYQ — JEE Advanced 2016 · Shift 2 · Q47 with Solution

View written solutionFree

Correct answer: C, D

Set up the motion in different regions

Let the magnetic-field region extend from $x=0$ to $x=3L$ , where $x$ denotes the position of the right edge of the square loop.

Since the loop has side $L$ , its left edge is at $x-L$ .

The magnetic flux through the loop changes only when the loop is partially inside the field region.

So the motion divides into four intervals:

$x<0$ : loop completely outside field
$0<x<L$ : loop entering field
$L<x<3L$ : loop completely inside field
$3L<x<4L$ : loop leaving field
$x>4L$ : loop completely outside again

Magnetic flux and induced current

The induced emf is

\mathcal E = -\frac{d\Phi}{dt}

and current is

I=\frac{\mathcal E}{R}.

We take counter-clockwise current as positive.

Region I: $x<0$

No part of loop is inside the field, so

\Phi=0,\quad I=0,\quad F=0.

Region II: $0<x<L$ (entering)

The overlap width is $x$ , so area inside field is

A=xL.

Hence flux into the page is

\Phi = B_0 xL.

Thus

\frac{d\Phi}{dt}=B_0L\frac{dx}{dt}=B_0Lv.

Therefore

\mathcal E=-B_0Lv.

With counter-clockwise positive, negative emf means clockwise current:

I=-\frac{B_0Lv}{R}.

Now force acts only on the vertical side inside the field (right edge). The magnetic force opposes the motion, so it is toward $-x$ .

Magnitude:

|F| = |I|LB_0 = \frac{B_0^2L^2}{R}v.

Hence

F_x=-\frac{B_0^2L^2}{R}v.

Equation of motion:

m\frac{dv}{dt}=-\frac{B_0^2L^2}{R}v.

Since $v=dx/dt$ ,

mv\frac{dv}{dx}=-\frac{B_0^2L^2}{R}v.

For $v\neq 0$ ,

\frac{dv}{dx}=-\frac{B_0^2L^2}{mR}=\text{constant}.

So in this interval, $v(x)$ decreases linearly with $x$ .

Since

I=-\frac{B_0L}{R}v,

$I(x)$ also varies linearly (negative), and

F(x)=-\frac{B_0^2L^2}{R}v

also varies linearly (negative).

Region III: $L<x<3L$ (fully inside)

Now the loop is completely inside the uniform field. Flux is constant:

\Phi=B_0L^2.

I=0,

and therefore

F=0.

Hence velocity remains constant in this region.

So $v(x)$ is a horizontal line here, equal to the value attained at $x=L$ .

Region IV: $3L<x<4L$ (leaving)

Now overlap width is $4L-x$ , so flux is

\Phi=B_0L(4L-x).

Then

\frac{d\Phi}{dt}=-B_0L\frac{dx}{dt}=-B_0Lv.

Thus

\mathcal E=-\frac{d\Phi}{dt}=+B_0Lv.

So current is counter-clockwise:

I=+\frac{B_0Lv}{R}.

Again the magnetic force opposes the motion, so

F_x=-\frac{B_0^2L^2}{R}v.

Therefore in this region also,

\frac{dv}{dx}=-\frac{B_0^2L^2}{mR},

so $v(x)$ again decreases linearly with $x$ .

Also,

I(x)=+\frac{B_0L}{R}v,

positive and linearly decreasing with $x$ , and

F(x)=-\frac{B_0^2L^2}{R}v,

negative and linearly decreasing in magnitude with $x$ .

Region V: $x>4L$

Loop is out of the field again:

I=0,\quad F=0,

and velocity becomes constant thereafter.

Qualitative shapes

From the above:

(i) Velocity $v(x)$

constant for $x<0$
linearly decreasing for $0<x<L$
constant for $L<x<3L$
linearly decreasing for $3L<x<4L$
constant for $x>4L$

So the correct $v(x)$ graph must have two downward linear segments separated by a flat segment.

(ii) Current $I(x)$

$0$ for $x<0$
negative during entry: $0<x<L$
$0$ for $L<x<3L$
positive during exit: $3L<x<4L$
$0$ for $x>4L$

Moreover, its magnitude is proportional to $v$ , so in each active interval it changes linearly with $x$ .

(iii) Force $F(x)$

$0$ for $x<0$
negative during entry
$0$ when fully inside
negative during exit
$0$ after leaving

Its magnitude is proportional to $v$ , so in each active interval it is a negative linear segment.

Match with options

The plots that match these behaviors are C and D.

So the correct choices are:

\boxed{C, D}

Comparison with stored answer

Stored correct answer: C, D

This matches the derived answer exactly.

More from Electromagnetic Induction

View written solutionFree

Correct answer: C, D

Set up the motion in different regions

Let the magnetic-field region extend from $x=0$ to $x=3L$ , where $x$ denotes the position of the right edge of the square loop.

Since the loop has side $L$ , its left edge is at $x-L$ .

The magnetic flux through the loop changes only when the loop is partially inside the field region.

So the motion divides into four intervals:

$x<0$ : loop completely outside field
$0<x<L$ : loop entering field
$L<x<3L$ : loop completely inside field
$3L<x<4L$ : loop leaving field
$x>4L$ : loop completely outside again

Magnetic flux and induced current

The induced emf is

\mathcal E = -\frac{d\Phi}{dt}

and current is

I=\frac{\mathcal E}{R}.

We take counter-clockwise current as positive.

Region I: $x<0$

No part of loop is inside the field, so

\Phi=0,\quad I=0,\quad F=0.

Region II: $0<x<L$ (entering)

The overlap width is $x$ , so area inside field is

A=xL.

Hence flux into the page is

\Phi = B_0 xL.

Thus

\frac{d\Phi}{dt}=B_0L\frac{dx}{dt}=B_0Lv.

Therefore

\mathcal E=-B_0Lv.

With counter-clockwise positive, negative emf means clockwise current:

I=-\frac{B_0Lv}{R}.

Now force acts only on the vertical side inside the field (right edge). The magnetic force opposes the motion, so it is toward $-x$ .

Magnitude:

|F| = |I|LB_0 = \frac{B_0^2L^2}{R}v.

Hence

F_x=-\frac{B_0^2L^2}{R}v.

Equation of motion:

m\frac{dv}{dt}=-\frac{B_0^2L^2}{R}v.

Since $v=dx/dt$ ,

mv\frac{dv}{dx}=-\frac{B_0^2L^2}{R}v.

For $v\neq 0$ ,

\frac{dv}{dx}=-\frac{B_0^2L^2}{mR}=\text{constant}.

So in this interval, $v(x)$ decreases linearly with $x$ .

Since

I=-\frac{B_0L}{R}v,

$I(x)$ also varies linearly (negative), and

F(x)=-\frac{B_0^2L^2}{R}v

also varies linearly (negative).

Region III: $L<x<3L$ (fully inside)

Now the loop is completely inside the uniform field. Flux is constant:

\Phi=B_0L^2.

I=0,

and therefore

F=0.

Hence velocity remains constant in this region.

So $v(x)$ is a horizontal line here, equal to the value attained at $x=L$ .

Region IV: $3L<x<4L$ (leaving)

Now overlap width is $4L-x$ , so flux is

\Phi=B_0L(4L-x).

Then

\frac{d\Phi}{dt}=-B_0L\frac{dx}{dt}=-B_0Lv.

Thus

\mathcal E=-\frac{d\Phi}{dt}=+B_0Lv.

So current is counter-clockwise:

I=+\frac{B_0Lv}{R}.

Again the magnetic force opposes the motion, so

F_x=-\frac{B_0^2L^2}{R}v.

Therefore in this region also,

\frac{dv}{dx}=-\frac{B_0^2L^2}{mR},

so $v(x)$ again decreases linearly with $x$ .

Also,

I(x)=+\frac{B_0L}{R}v,

positive and linearly decreasing with $x$ , and

F(x)=-\frac{B_0^2L^2}{R}v,

negative and linearly decreasing in magnitude with $x$ .

Region V: $x>4L$

Loop is out of the field again:

I=0,\quad F=0,

and velocity becomes constant thereafter.

Qualitative shapes

From the above:

(i) Velocity $v(x)$

constant for $x<0$
linearly decreasing for $0<x<L$
constant for $L<x<3L$
linearly decreasing for $3L<x<4L$
constant for $x>4L$

So the correct $v(x)$ graph must have two downward linear segments separated by a flat segment.

(ii) Current $I(x)$

$0$ for $x<0$
negative during entry: $0<x<L$
$0$ for $L<x<3L$
positive during exit: $3L<x<4L$
$0$ for $x>4L$

Moreover, its magnitude is proportional to $v$ , so in each active interval it changes linearly with $x$ .

(iii) Force $F(x)$

$0$ for $x<0$
negative during entry
$0$ when fully inside
negative during exit
$0$ after leaving

Its magnitude is proportional to $v$ , so in each active interval it is a negative linear segment.

Match with options

The plots that match these behaviors are C and D.

So the correct choices are:

\boxed{C, D}

Comparison with stored answer

Stored correct answer: C, D

This matches the derived answer exactly.

Electromagnetic Induction question

Electromagnetic Induction question

Region I: $x<0$

Region II: $0<x<L$ (entering)

Region III: $L<x<3L$ (fully inside)

Region IV: $3L<x<4L$ (leaving)

Region V: $x>4L$

(i) Velocity $v(x)$

(ii) Current $I(x)$

(iii) Force $F(x)$

More from Electromagnetic Induction

Electromagnetic Induction question

Electromagnetic Induction question

Region I: $x<0$

Region II: $0<x<L$ (entering)

Region III: $L<x<3L$ (fully inside)

Region IV: $3L<x<4L$ (leaving)

Region V: $x>4L$

(i) Velocity $v(x)$

(ii) Current $I(x)$

(iii) Force $F(x)$

More from Electromagnetic Induction

Region I: x<0x<0x<0

Region II: 0<x<L0<x<L0<x<L (entering)

Region III: L<x<3LL<x<3LL<x<3L (fully inside)

Region IV: 3L<x<4L3L<x<4L3L<x<4L (leaving)

Region V: x>4Lx>4Lx>4L

(i) Velocity v(x)v(x)v(x)

(ii) Current I(x)I(x)I(x)

(iii) Force F(x)F(x)F(x)

More from Electromagnetic Induction

Region I: x<0x<0x<0

Region II: 0<x<L0<x<L0<x<L (entering)

Region III: L<x<3LL<x<3LL<x<3L (fully inside)

Region IV: 3L<x<4L3L<x<4L3L<x<4L (leaving)

Region V: x>4Lx>4Lx>4L

(i) Velocity v(x)v(x)v(x)

(ii) Current I(x)I(x)I(x)

(iii) Force F(x)F(x)F(x)

More from Electromagnetic Induction

Region I: $x<0$

Region II: $0<x<L$ (entering)

Region III: $L<x<3L$ (fully inside)

Region IV: $3L<x<4L$ (leaving)

Region V: $x>4L$

(i) Velocity $v(x)$

(ii) Current $I(x)$

(iii) Force $F(x)$

Region I: $x<0$

Region II: $0<x<L$ (entering)

Region III: $L<x<3L$ (fully inside)

Region IV: $3L<x<4L$ (leaving)

Region V: $x>4L$

(i) Velocity $v(x)$

(ii) Current $I(x)$

(iii) Force $F(x)$